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Paper of JEE exam which was held on 1st Feb 2024 2nd session
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FINAL JEE–MAIN EXAMINATION – JANUARY, 2024(Held On Thursday 01st February, 2024) TIME : 3 : 00 PM to 06 : 00 PM
MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A
Sol. px2 + qx – r = 0
2
1. Let f(x) = |2x +5|x|–3|,xR. If m and n denote the
p = A, q = AR, r = AR2
number of points where f is not continuous and not
Ax2 + ARx – AR2 = 0
differentiable respectively, then m + n is equal to :
(1) 5 (2) 2 x2 + Rx – R2 = 0
(3) 0 (4) 3 1 1 3
Ans. (4) 4
Sol. f(x) = |2x2+5|x|–3| 3 R 4
R
Graph of y = |2x2 + 5x – 3| 4 R 2 4 3
16
()2 = ()2 – 4 = R2 – 4(–R2) = 5
9
= 80/9
3. The number of solutions of the equation 4 sin2x – 4
cos3x + 9 – 4 cosx = 0; x [ –2, 2] is :
–3 –5/4 1/2 (1) 1
(2) 3
Graph of f(x) (3) 2
(4) 0
Ans. (4)
Sol. 4sin2x – 4cos3x + 9 – 4 cosx = 0 ; x [ – 2, 2]
4 – 4cos2x – 4cos3x + 9 – 4 cosx = 0
4cos3x + 4cos2x + 4 cosx – 13 = 0
4cos3x + 4cos2x + 4cosx = 13
0
L.H.S. 12 can’t be equal to 13.
1
1
4. The value of 0
(2x 3 3x x 2
1) 3 dx is equal to:
Number of points of discontinuity = 0 = m
(1) 0
Number of points of non-differentiability = 3 = n
(2) 1
2. Let and be the roots of the equation px2 + qx – (3) 2
r = 0, where p 0. If p, q and r be the consecutive (4) –1
1 1 3 Ans. (1)
terms of a non-constant G.P and , then
4 1
1 3
3 2
the value of (– is : Sol. I 2x 3x x 1 dx
0
80
(1) (2) 9 2a
9 Using f x dx where f(2a–x) = –f(x)
0
20
(3) (4) 8 Here f(1–x) = f(x)
3
I=0
Ans. (1)
, 18
x 2 y2 1 2
5. Let P be a point on the ellipse 1 . Let the
9 4 x3 x 3
Sol. 3
line passing through P and parallel to y-axis meet
the circle x2 + y2 = 9 at point Q such that P and Q 12 6
are on the same side of the x-axis. Then, the 1 2
x3 x3 18 1 1
eccentricity of the locus of the point R on PQ such t 7 18c6 2 c6 12 . 26
3 3
that PR : RQ = 4 : 3 as P moves on the ellipse, is :
11 13 6 12
(1) (2) 1 2
19 21 x3 x3 1 1
t13 18c12 2
18
c12 . 12 .x –6
3 2
6
139 13 3
(3) (4)
23 7
Ans. (4) m 18c6 .3–12.2 –6 : n 18c12 .2–12.3–6
1 1
2
n 3 2 .3 3 3
–12 –6
9
Q x2 y2 12 6
P 1 m 3 .2 2 4
9 4
P(3cos , 2sin ) 7. Let be a non-zero real number. Suppose f : R
Q(3cos , 3sin) R is a differentiable function such that f (0) = 2 and
lim f x 1 . If f '(x) = f(x) +3, for all x R,
x
Sol.
4 3 then f (–loge2) is equal to____.
(1) 3 (2) 5
P R Q (3) 9 (4) 7
(3C, 2S) (h, k) (3C, 3S) Ans. (3 OR BONUS)
h = 3cos;
Sol. f(0) = 2, lim f x 1
x
18
k sin f’(x) – x.f(x) = 3
7
I.F = e–x
x 2 49y 2
locus = 1 y(e–x) = 3.e x dx
9 324
324 117 13 3e x
e 1 f(x). (e–x) = c
49 9 21 7
6. Let m and n be the coefficients of seventh and 3 3
x=0 2 c c2 (1)
thirteenth terms respectively in the expansion of
18 3
1 1
f(x) = c.ex
1 x3 1 . Then n 3
3 2 is :
m 3
2x 3 x – 1 = c(0)
4 1
(1) (2) = –3 c = 1
9 9
3
1 9 f ( ln 2) c.ex
(3) (4)
4 4
= 1 + e3ln2 =9
Ans. (4)
(But should be greater than 0 for finite
value of c)
, 8. Let P and Q be the points on the line A (1,3,2)
x 3 y 4 z 1
which are at a distance of 6
8 2 2
units from the point R (1,2,3). If the centroid of the
1: 1
triangle PQR is (,,), then 2+2 +2 is:
B D C (3, 6, 7)
(1) 26 (–2, 8, 0) 1 7
2 , 7, 2
(2) 36 Sol.
(3) 18 A(1, 3, 2); B(–2, 8, 0); C(3, 6, 7);
(4) 24 AC 2iˆ 3jˆ 5kˆ
Ans. (3)
AB = 9 25 4 38
Sol.
R(1,2,3) AC = 4 9 25 38
1 3 1
AD ˆi 4jˆ kˆ (iˆ 8jˆ 3k)
ˆ
2 2 2
Length of projection of AD on AC
P Q
AD.AC 37
P(8 – 3, 2 + 4, 2 – 1) =
| AC | 2 38
PR = 6
10. Let Sn denote the sum of the first n terms of an
(8 – 4)2 + ( 2 + 2)2 + (2 – 4)2 = 36
arithmetic progression. If S10 = 390 and the ratio of
= 0, 1 the tenth and the fifth terms is 15 : 7, then S15 –S5
Hence P(–3, 4, –1) & Q(5, 6, 1) is equal to:
Centroid of PQR = (1, 4, 1) (,, ) (1) 800
+ + = 18 (2) 890
9. Consider a ABC where A(1,2,3,), B(–2,8,0) and (3) 790
C(3,6,7). If the angle bisector of BAC meets the (4) 690
line BC at D, then the length of the projection of Ans. (3)
Sol. S10 = 390
the vector AD on the vector AC is:
10
2a 10 1 d 390
(1)
37 2
2 38
2a + 9d = 78 (1)
38 t10 15 a 9d 15
(2) 8a 3d (2)
2 t5 7 a 4d 7
39
(3) From (1) & (2) a=3&d=8
2 38
15 5
(4) 19
S15 – S5 6 14 8 6 4 8
2 2
Ans. (1) 15 118 5 38
=790
2
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